Lesson Video: Kinetic Friction Physics • 9th Grade

In this video, we will learn how to analyze the forces acting on objects on frictional surfaces when they move along these surfaces.

18:04

Video Transcript

In this video, we’re looking at kinetic friction. Kinetic friction describes the frictional force between two surfaces that are moving past each other. For example, here, kinetic friction produced by the skis sliding over the ski slope creates a force that opposes the skier’s motion. In this video, we’ll look at how we can calculate the size of the kinetic frictional force between two surfaces using the coefficient of kinetic friction 𝜇 k.

When we’re learning about friction, we usually start with static friction, that is, the frictional force that acts between stationary objects. For example, if we consider this person who’s trying and failing to push a heavy object that on the ground, it’s static friction between the object and the ground that prevents the object from moving. Now, static friction always acts in such a way that it balances out any applied forces and prevents an object from moving.

We can draw force vectors on our diagram to represent this happening. The person pushing the books produces an applied force, which we can call 𝐹 A, acting to the right, and the static frictional force 𝐹 F acts in the opposite direction. We also have the object’s weight acting downwards, which is equal to its mass multiplied by gravitational acceleration 𝑔 and a surface force or normal force 𝐹 N acting upwards.

Now, as long as this object remains stationary, we know that the magnitude of the frictional force must be equal to the magnitude of the applied force. If this person pushes the box harder, increasing the applied force 𝐹 A, then 𝐹 F increases as well and the forces stay balanced. If the forces are balanced, that means that the resulting force is zero and the object remains stationary. However, a static frictional force can only get so big. In fact, we know that it takes a maximum value of 𝐹 N multiplied by 𝜇 S, that is, the normal reaction force multiplied by the coefficient of static friction. So we can say that the magnitude of the frictional force 𝐹 F is less than or equal to the magnitude of this quantity.

What this means is that a frictional force will act to perfectly counteract any applied force that’s preventing an object from moving up to a maximum magnitude given by 𝐹 N times 𝜇 S. So let’s say that we increase our applied force up to this maximum value. So now, the magnitude of our applied force is equal to the magnitude of 𝐹 N multiplied by 𝜇 S. When we do this, we see that 𝐹 F increases to the same magnitude. So we can say that the magnitude of 𝐹 F is also equal to the magnitude of 𝐹 N times 𝜇 S. That’s balancing out the applied force.

This represents the point where the frictional force is at the biggest value it can possibly take. This means that even the tiniest additional increase in the applied force, so that the applied force is now bigger than 𝐹 N times 𝜇 S, means that the horizontal forces are no longer balanced. And the object starts to accelerate to the right. In other words, the object starts to slip across the floor.

Now, as soon as this starts to happen, static friction no longer applies. We’re now talking about kinetic friction. Now, as soon as an object starts to slip, we actually find that the frictional force acting on it decreases. To describe this mathematically, we say that for sliding objects, the frictional force 𝐹 F is equal to the normal reaction force 𝐹 N multiplied by a different coefficient of friction, specifically the coefficient of kinetic friction 𝜇 K. This equation is often written with a negative sign just to signify the fact that the frictional force once again opposes motion.

The coefficient of kinetic friction 𝜇 K has several similarities to the coefficient of static friction. They are both dimensionless numbers which we multiply by the normal reaction force to calculate the size of a frictional force. They also both take different values for different materials. We usually find that these values are between zero and one. Importantly, the coefficient of kinetic friction is generally smaller than the coefficient of static friction, which means that the frictional force experienced by a moving object is smaller than the maximum static friction that it could experience. Unlike static friction, which can take any value from zero up to the magnitude of 𝐹 N times 𝜇 S, kinetic friction takes a constant value regardless of the applied force.

It’s worth noting that this isn’t a completely accurate model. And in real-life situations, we find that the frictional force experienced by a moving object varies in a complicated way depending on how quickly it’s traveling, as well as other properties of the surfaces. But this formula gives us a really simple way of describing kinetic friction that still describes the main behavior that we see with real-life objects. Notice that the kinetic frictional force is proportional to the normal reaction force experienced by the object.

In this example, where the object is on a horizontal surface, we can see that the normal reaction force acts in the opposite direction to the weight of the object 𝑚𝑔. Since the normal reaction force and the weight of the object are the only two forces acting in the vertical direction and the object isn’t moving in the vertical direction, we know that these two forces must be in equilibrium. In other words, the magnitude of the normal reaction force 𝐹 N is equal to the magnitude of the weight of the object 𝑚𝑔. And this is the same for any object on a horizontal surface as long as there are no other forces with vertical components acting on the object.

So in situations like this one, where the object is on a horizontal surface and the only applied force is acting horizontally, we can say that the kinetic friction force is equal to negative 𝑚𝑔 times 𝜇 K. However, when we consider an object on an incline, we can see that the normal reaction force, which is always perpendicular to the surface, and the weight force, which always acts vertically downward, no longer directly oppose each other, which means we can no longer say that they’re exactly the same size. So if we want to calculate the kinetic friction acting on an inclined surface, we need to be careful to first calculate the normal reaction force.

One final thing to consider is that when an object slides across a surface, kinetic friction acts to slow the object down, thereby removing energy from the system. Specifically, kinetic friction takes energy away from the moving object and dissipates it in the form of heat. Now, when we put energy into an object, we can say that we’re doing work on that object. So because kinetic friction is taking energy out of an object, we can say that it does a negative amount of work on the object. Generally, we calculate the work done by a force by multiplying the size of that force by the distance traveled in the direction of that force. For example, if this person manages to move the object a distance 𝑑 by applying a constant force 𝐹 A, then the work that they do on the box is equal to 𝐹 A times 𝑑.

Now we can also use this formula to calculate the work done by friction. And to do this, we’d multiply the frictional force 𝐹 F by the distance moved. We should note that this formula tells us the work done when the force and the distance moved are in the same direction. But when we’re calculating the work done by friction, let’s call this 𝑊 F, the frictional force and the distance traveled are in opposite directions. To signify this, we say that the frictional force 𝐹 F is negative, and we can see that this would result in a negative value of work done. Signifying the kinetic friction takes energy away from the object.

We could equivalently say that some positive amount of energy is dissipated by friction. And if we describe it in this way, then we don’t need to worry about the negative sign. We can say that the energy dissipated by friction 𝐸 is equal to the frictional force 𝐹 F times the distance traveled 𝑑. So now that we’ve seen how we can calculate the size of the force generated by kinetic friction and the amount of energy that this force dissipates, let’s look at a question that puts some of these ideas into practice.

A stone with a mass of 8.2 kilograms is initially at rest on a rough surface. The coefficient of static friction between the stone and the surface is 0.70, and the coefficient of kinetic friction between the stone and the surface is 0.60. A force is applied to the stone parallel to the surface that negligibly exceeds the force required to make the stone move across the surface. What is the magnitude of the acceleration of the stone across the surface? (A) 0.98 meters per second squared, (B) 3.4 meters per second squared, (C) 4.1 meters per second squared, (D) 5.9 meters per second squared, or (E) 6.9 meters per second squared.

So in this question, we’re considering a stone on a surface. Because we haven’t been told otherwise, let’s assume that the surface is horizontal. So here’s our stone and here’s the surface. Now, one of the first things we can notice about this question is that we’re being asked to find acceleration and we’ve been given information about forces such as an applied force and frictional forces. Newton’s second law tells us that the net force acting on an object, 𝐹 net, is equal to the mass of that object, 𝑚, multiplied by the acceleration of the object, 𝑎.

If we divide both sides of this equation by 𝑚, we’re left with 𝐹 net over 𝑚 equals 𝑎, showing us that if we can find the net force acting on an object, all we need to do is divide this by the mass and we’ll obtain the acceleration of the object. In order to work out the net force that’s acting on the stone in this question, we need to think about all of the forces that are acting on it. So let’s start by drawing these forces onto a diagram. Firstly, we know that the stone’s weight equal to its mass multiplied by gravitational acceleration acts downwards. And we know that the surface exerts a normal reaction force 𝐹 N on the stone.

Since the surface in this case is horizontal, we know that the normal reaction force will be vertically upward. We’re also told in the question that a force is applied to the stone parallel to the surface. Let’s call this applied force 𝐹 A. And crucially, we’re told in this question that the surface is rough. This is just a way of saying that friction acts between the object and the surface. We know that frictional forces act parallel to the surface and oppose motion. So we can draw our frictional force 𝐹 F acting to the left. So now we can see all of the forces acting on the stone.

One of the first things we can notice is that all of these forces either act horizontally or vertically. This makes our job easier as we can consider vertical and horizontal motions separately without needing to resolve any of our forces into their components. Looking at the vertical forces, we can conclude that the normal reaction force and the weight must be equal in magnitude. Since the stone doesn’t move in the vertical direction, the vertical forces must be balanced. In other words, the magnitude of 𝐹 N is equal to the magnitude of 𝑚𝑔. And this is something that will come in handy later in the question.

Now we can look at the horizontal forces. We’re told in the question that the applied force negligibly exceeds the force required to make the stone move. This tells us that the magnitude of the applied force must be greater than the magnitude of the frictional force, resulting in a net force acting to the right, thus causing the stone to accelerate to the right. In order to find this acceleration, we need to find the net horizontal force. And to do this, we need to calculate the size of the applied force and the frictional force.

Let’s start with the applied force. The question tells us that this applied force negligibly exceeds the force required to make the stone move. So what exactly does this mean? Let’s recall that an object on a rough surface will be prevented from moving by static friction. Static friction always acts to oppose an applied force that would otherwise cause an object to accelerate. In cases where static friction is acting, we can say that 𝐹 F, the frictional force, is equal to negative 𝐹 A. That is, the frictional force has the same magnitude as the applied force, but the negative sign shows us that it acts in the opposite direction, thus opposing motion.

Now, static friction can only produce a force up to a certain maximum value. This maximum value is given by 𝐹 N, the normal reaction force, multiplied by 𝜇 S, the coefficient of static friction. So we can say that the magnitude of the static frictional force 𝐹 F is always less than or equal to the magnitude of 𝐹 N times 𝜇 S. If we have a stone that’s initially at rest and then we apply a small force, then we’ll find that a static frictional force acting in the opposite direction will balance it out. If we slightly increase the size of the applied force, then the size of the static frictional force increases too, preventing the stone from moving.

However, we know that the static frictional force can never exceed this value. So if we apply a force that’s greater than this, then static friction will not be able to counteract it. This means that the forces become imbalanced, thereby producing a net force in the direction of the applied force, causing the object to accelerate. So the magnitude of applied force required to make the stone accelerate is anything larger than this. In fact, the question tells us that the force negligibly exceeds this value. To help us see what this means, let’s work out the numerical value of 𝐹 N times 𝜇 S.

First, we can write down that the maximum magnitude of the static friction is equal to 𝐹 N times 𝜇 S. We’re told in the question that 𝜇 S is equal to 0.7, and we’ve shown that 𝐹 N is equal in magnitude to 𝑚𝑔. This means that the maximum magnitude of the static friction is equal to 0.70 times 𝑚𝑔. We’re told that the mass of the stone is 8.2 kilograms, and we know the value of 𝑔 is 9.8 meters per second squared. So the maximum magnitude of the static frictional force is equal to 0.70 times 8.2 kilograms times 9.8 meters per second squared, which is equal to 56.252 newtons.

So given that this is the maximum size of the static frictional force, we know that if the magnitude of the applied force exceeds this by even a negligible amount, then the stone will accelerate. So, for example, this could be achieved by an applied force with a magnitude of 56.253 newtons or even by an applied force with a magnitude of 56.252001 newtons.

In fact, if the magnitude of the applied force negligibly exceeds this value, then if we write the magnitude of the applied force to the same number of significant figures as the magnitude of the frictional force, then they’re written as exactly the same number. So for the purposes of our calculation, we can say that the magnitude of the applied force required to make the stone move is equal to the maximum magnitude of the static frictional force. And we’ve shown that this is equal to 𝑚𝑔 times 𝜇 S.

Okay, so now we found the size of the applied force, we need to find out the size of the frictional force. Now, the question states that the stone moves across the surface. So it’s important to remember that as soon as an object starts to slide along a surface, the rules of static friction no longer apply. Instead, we’re dealing with kinetic friction. Now, unlike a static frictional force, which can vary with the applied force, a kinetic frictional force is constant. Its value is given by the normal reaction force multiplied by the kinetic coefficient of friction 𝜇 K. And this equation is often written with a negative sign to show that the frictional force opposes motion.

We can see that this quantity is similar to the maximum static friction. We’re just multiplying the normal force by a different coefficient of friction. In the question, we’re told that the coefficient of kinetic friction is 0.60, and we can note that this is lower than the coefficient of static friction. In fact, this is the case for all objects on all surfaces. As soon as an object starts to slip across a surface, its coefficient of friction effectively drops.

This means that if we have an object resting on a rough surface and we gradually increase the applied force until it slips, we find that at the moment it slips, the frictional force decreases. This means that the forces acting on a stone might look like this just before it slips. But at the moment it begins to slip, the frictional force decreases. And the magnitude of this new frictional force is given by the normal force multiplied by the coefficient of kinetic friction.

Since in this case we’ve shown that the normal force is equal in magnitude to the weight of the stone, we know that the frictional force is given by negative 𝑚𝑔 times 𝜇 K. Remember that the negative sign in this formula only exists to remind us that the frictional force opposes motion, so we can remove it from our expression. Now that we have expressions for the horizontal forces acting on the stone, we’re now ready to apply Newton’s second law to calculate the acceleration.

Using the common convention that forces acting to the right are positive and forces acting to the left are negative, the net force acting on the stone is equal to the magnitude of 𝐹 A minus the magnitude of 𝐹 F. And we’ve shown that 𝐹 A has a magnitude of 𝑚𝑔 times 𝜇 S and 𝐹 F has a magnitude of 𝑚𝑔 times 𝜇 K. Newton’s second law tells us the acceleration is given by the net force, that’s this expression, divided by the mass. So the acceleration of the stone is given by this expression. We can take out a factor of 𝑚𝑔 in the numerator, giving us 𝑚𝑔 times 𝜇 S minus 𝜇 K over 𝑚. And we can then cancel the common factor of 𝑚 in the numerator and denominator, leaving us with 𝑔 times 𝜇 S minus 𝜇 K.

Substituting in the values of 𝑔, 𝜇 S, and 𝜇 K gives us 9.8 meters per second squared multiplied by 0.70 minus 0.60. 0.70 minus 0.60 is 0.10. And multiplying these numbers together gives us 0.98 meters per second squared. So the correct answer to this question is option (A). It’s worth noting that, in this question, since the stone is moving, static friction isn’t actually acting on it. But we still need to think about static friction in order to figure out the applied force which is required to make it move.

With this example finished, now let’s recap the key points that we’ve looked at in this video. We’ve seen that kinetic friction is a contact force that opposes the motion of objects moving across rough surfaces. The size of this force depends on two things: the normal reaction force 𝐹 N and the coefficient of kinetic friction 𝜇 K. And 𝜇 K itself depends on the materials involved. The value of the kinetic frictional force is given by negative 𝐹 N times 𝜇 K, where the negative sign just signifies that this frictional force opposes motion. And finally, we’ve seen that the energy 𝐸 dissipated by a frictional force is equal to the frictional force multiplied by the distance the object moves.

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